Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $G$ be a finite group acting on a topological space $X$. (In my applications, $X$ is the classifying space of a compact Lie group.) Let $H^\ast(-)=H^\ast(-;\mathbb{Z}/2\mathbb{Z})$ denote mod 2 singular cohomology.

By the work of Quillen ("The Spectrum of an Equivariant Cohomology Ring: I"), if $H^\ast(X)$ is a finitely generated $\mathbb{Z}/2\mathbb{Z}$-module, then the equivariant cohomology $$H^\ast_G(X) = H^\ast(EG\times_G X)$$ is a finitely generated as a $\mathbb{Z}/2\mathbb{Z}$-algebra.

My question is, if we only assume that $H^\ast(X)$ is finitely generated as an algebra, does the conclusion still hold, ie is the equivariant cohomology $H^\ast_G(X)$ necessarily finitely generated as an algebra?

More background

The standard tool for computing equivariant cohomology is the Leray-Serre spectral sequence of the fibration $X\to EG\times_G X \to BG$, which has $$H^\ast(BG,H^\ast(X)) \Longrightarrow H^\ast(EG\times_G X)$$ as algebras.

Now if $G$ acts trivially on $H^\ast(X)$ then the $E_2$-page can be identified with $$H^\ast(BG)\otimes H^\ast(X)$$ as algebras. By a classical result of Evens, the mod 2 cohomology algebra of a finite group is finitely generated. Since the tensor product of finitely generated algebras is again finitely generated, so is this $E_2$-page, and hence so is the equivariant cohomology algebra. (see the comments made by Algori and Ralph below.)

However, in general the coefficients in the $E_2$-page are twisted by the action of $G$ on $H^\ast(X)$ (which in my case happens to be non-trivial), and so I can't see how the argument would go.

At the other extreme, if $G$ acts freely on $X$ then we are basically asking if there is a finite covering space $X\to Y$ such that $H^\ast(X)$ is finitely generated as an algebra but $H^\ast(Y)$ is not. Such a beast would give a counter-example.

(I'm also rather sure that if the answer to my question was yes, Quillen would have told us so!)

share|improve this question
2  
Mark -- re "and hence so is the equivariant cohomology algebra": $E_3$ is the quotient of a subalgebra of $E_2$; now, a subalgebra of a finitely generated algebra is not necessarily finitely generated, so I think an extra argument is needed here. –  algori Oct 25 '11 at 20:16
1  
$E_2$ is always a finitely generated algebra over $k = \mathbb{F}_2$: By Noether normalization $H^\ast(X)$ is a finitely generated module over a polynomial ring $k[x_1,...x_n]$ and wlog we may assume that the $x_i$ have the same degree. By Dickson invariants, $k[x_1,...,x_n]$ is a finitely generated module over the $k$-polynomial ring $R := k[x_1,...,x_n]^G$. Thus $H^\ast(X)$ is a finitely generated $R$-module. Now by Evens' theorem $H^\ast(BG;R)$ is a finitely generated $R$-algebra and $E_2=H^\ast(BG;H^\ast(X))$ is a finitely generated $H^\ast(BG;R)$-module. –  Ralph Oct 26 '11 at 23:21
1  
... Summarizing, $E_2$ is a finitely generated module over a finitely generated $k$-algebra. In particular, $E_2$ is a finitely generated $k$-algebra. --- But as pointed out by algori (and can be seen in Evens' original proof) this doesn't suffice to show finite generation. –  Ralph Oct 26 '11 at 23:21
    
@Ralph: Is $E_2$ a finitely generated $H^\ast(BG)$-module? If it were, I think this would answer the question as the ring $H^\ast(BG)$ is commutative and Noetherian, and so subquotients of f.g. modules are f.g. –  Mark Grant Oct 27 '11 at 13:09
2  
In this case $H_G^\ast(X;k)$ is a finitely generated $k$-algebra: By a theorem of Nakaoka, $H^\ast(EG \times_G Y^r;k) \cong H^\ast(BG;H^\ast(Y;k)^{\otimes r})$ as $k$-algebras and the latter is isomophic to $H^\ast(BG;H^\ast(X;k))$ which is just the $E_2$-term of the spectral sequence discussed above. From the considerations in my first comment now the desired result follows. --- You can find Nakaoka's theorem in arxiv.org/abs/0711.5017, Theorem 2.1. Note: $X^{\otimes \Omega}$ in that theorem is a typo, it should be $X^\Omega$ as is clear from context or from the proof. –  Ralph Oct 27 '11 at 22:48

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.